Question

if Sn denote the sum of n term of an ap whose common difference is d and first term is a , find Sn-2Sn-1+Sn+2

Answer 1

We know Sum of n terms of an AP

                                Sn = n/2[2a+(n-1)d]

Similarly we can say that Sum of n-1 Terms can be written as

                               Sn-1 = n-1/2[2a+(n-1-1)d]
                               Sn-1 = n-1/2[2a+(n-2)d]
and also
   Sum of n-2 terms of AP
          
                               Sn-2 = n-2/2[2a+(n-2-1)d]
                               Sn-2 = n-2/2[2a+(n-3)d]

Now we have to find Sn-2Sn-1+Sn-2

Thus putting the above equations

Sn-2Sn-1+Sn-2 = n/2[2a+(n-1)d] - 2 *n-1/2[2a+(n-2)d] + n-2/2[2+(n-3)d]
                          = n/2[2a+(n-1)d] - {(n-1)[2a+(n-2)d]} + n/2[2a+(n-3)d] -                                          2/2[2a+(n-3)d]
                            = n/2[4a+2nd-4d] - [2an-2a+n^2d-nd-2nd+2d] - 2a-nd+3d
                             = d

Thus value of Sn-2Sn-1+Sn-2 is equal to d

Answer 2

Answer:

Given : a is first term and d be the common difference.

↠ Sn = (n/2)[ 2a + ( n -1) d] Now Sn - 2Sn-1 + Sn + 2

↠ (n/2)[ 2a + ( n -1) d] - 2(n - 1)/2)[ 2a + ( n - 1 -1) d] + (n +2) / 2)[ 2a + ( n + 2 -1) d]

↠ (1/2)[ 2an + n( n -1) d] + [ 4a(n - 1) + 2(n - 1)( n - 2) d] +[ 2a(n + 2)+ ( n + 1) (n + 2) d]

↠ (1/2)[ 2a[ n - 2n + 2 + n + 2] + d [ n2 - n - 2n2 + 6n - 4 + n2 + 3n + 2] ]

↠ (1/2)[ 2a(4) + d(8n - 2) ]

↠ 4a + (4n - 1)d